Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{-3r - 9}{r^2 - 8r + 16} \times \dfrac{7r - 28}{r + 3} $
Solution: First factor the quadratic. $q = \dfrac{-3r - 9}{(r - 4)(r - 4)} \times \dfrac{7r - 28}{r + 3} $ Then factor out any other terms. $q = \dfrac{-3(r + 3)}{(r - 4)(r - 4)} \times \dfrac{7(r - 4)}{r + 3} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -3(r + 3) \times 7(r - 4) } { (r - 4)(r - 4) \times (r + 3) } $ $q = \dfrac{ -21(r + 3)(r - 4)}{ (r - 4)(r - 4)(r + 3)} $ Notice that $(r + 3)$ and $(r - 4)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -21(r + 3)\cancel{(r - 4)}}{ \cancel{(r - 4)}(r - 4)(r + 3)} $ We are dividing by $r - 4$ , so $r - 4 \neq 0$ Therefore, $r \neq 4$ $q = \dfrac{ -21\cancel{(r + 3)}\cancel{(r - 4)}}{ \cancel{(r - 4)}(r - 4)\cancel{(r + 3)}} $ We are dividing by $r + 3$ , so $r + 3 \neq 0$ Therefore, $r \neq -3$ $q = \dfrac{-21}{r - 4} ; \space r \neq 4 ; \space r \neq -3 $